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Â >> Hi, and welcome back to mechanics of materials part one.

Â In the next few modules we're going to be going through a bit of mathematics

Â to develop some very important relationships.

Â The beauty of these recordings is that even though I'll go rather quickly,

Â you can go back and stop and

Â start them to make sure that you understand every step in the process.

Â So today's learning outcome is to find the maximum and

Â minimum in-plane principal stresses.

Â And so, here's where we left off last time.

Â We have transformation equation for stress where we know stress on a,

Â 0:46

in certain directions, and we can find the normal and

Â the shear stresses on any other plane using this transformation equations.

Â And we found an angle to what we call, where we find the maximum or

Â minimum normal stresses to occur with this tangent relationship

Â where theta sub P is the angle to which is defying the principle plains.

Â And notice that on the block itself, we're working with theta sub P.

Â In the formula, we're working with 2 theta sub P.

Â And so there's that relationship.

Â And so let's look at this graphically.

Â Since it's a tangent, opposite would be tau xy over adjacent,

Â which would be sigma sub x minus sigma sub y over two.

Â Knowing those two perpendicular sides by the Pythagorean Theorem,

Â we can find the hypotenuse.

Â And so let's now consider both tau xy and

Â sigma x, and sigma y, sigma x minus sigma y over two being the same sign.

Â So lets first consider them to both be positive as I've shown here.

Â And if they're going to both be positive,

Â we see that 2 theta sub P has to be between 0 and 90 degrees.

Â They can also have the same sign if they're both negative,

Â on the top and the bottom.

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And then we're going to be between 180 degrees and

Â 270 degrees because we have a negative in the tau xy, or the shear stress direction,

Â and a negative in the sigma x minus sigma y over 2 direction.

Â And so, there are also two values of theta sub P on the stress block itself,

Â and they're going to be one half the magnitude of the 2 theta sub P.

Â So, the theta sub P's, the angles to the principal planes,

Â will be 0 to 45 degrees, and half of this, which is 90 degrees or greater.

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these two values, tau xy and

Â sigma sub x minus sigma sub y over 2 have, end up being a negative sign.

Â You could have a negative tau xy in the numerator, and

Â a positive sigma sub x minus sigma sub y over 2 value in the denominator.

Â Or you could have a positive tao sub xy value in the numerator and

Â a negative sigma sub x minus sigma sub y over 2 value in the denominator.

Â And if that occurs, tangent of 2 theta sub pi is going to be negative.

Â And so, therefore, we're going to have the,

Â one would be negative, the other one would be positive.

Â And so, we would be somewhere between 0 and

Â minus 90 degrees, or somewhere between minus

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And now we can take our strain transformation, excuse me,

Â stress transformation equation for sigma sub n, the normal stress,

Â and we can substitute in cosine 2 theta sub P.

Â So this should be theta sub P now.

Â And sine 2 theta sub P, which will end up being theta sub P now.

Â So these, now we're talking about specific angles,

Â so these will be theta sub P and theta sub P.

Â And this is the result I get by substituting those in.

Â Okay.

Â Plus or minus.

Â We can, this value and this value is the same, so we can factor that out.

Â And we have these two terms times this entire value here.

Â We see that we have sigma sub x minus sigma sub y over 2 squared,

Â and then the square root of that.

Â So that leaves a square root in the numerator.

Â Okay?

Â And so here is our expression boiled down.

Â And that now gives us the normal stresses on these principal planes.

Â We have two of them, and

Â I've labeled them principal stress number one and principal stress number two.

Â Here it is again.

Â So, at some angle, theta, we rotate our block.

Â And at that angle theta we end up with what's called principal stresses.

Â And I've shown them.

Â They can be both positive.

Â They can be both negative.

Â One could be positive, one could be negative.

Â But you'll note in this algebraic,

Â in this development, I've considered max and mins to be algebraic quantities.

Â And so, as I said, they could both be positive.

Â I might have plus 1,500 and plus 500.

Â Or I might have plus 800 and minus 200.

Â Or I might have minus 400 and minus 1700.

Â But in our calculations for

Â our engineering problems, when we use the term maximum we're going to refer to

Â the stresses with the largest absolute value, or the largest magnitude.

Â And so, for example, I have sigma sub 1, maybe a 700 megapascals.

Â Sigma sub 2 is minus 1200 megapascals.

Â This one is in tension.

Â This one is in compression.

Â But if I'm talking about the maximum normal stress, I'm going to refer to

Â the one that is the maximum absolute value when we do engineering problems.

Â Okay.

Â Here's our result again.

Â We'll see from this that we can end up with what we call a stress invariant.

Â If I add these two sigma sub 1s and sigma sub 2s together,

Â I get sigma sub 1 plus sigma sub 2 on the left hand side.

Â On the right hand side, this plus and minus will cancel out this part.

Â And so, I'll have sigma sub x plus sigma sub y over two plus sigma sub x plus

Â sigma sub y over 2, or sigma sub x plus sigma sub y.

Â That's a very important result.

Â It's the stress invariance.

Â So what it's saying is, on any two orthogonal planes

Â the sum of the normal stresses are going to be constant.

Â And so, no matter how we turn the block, the sum of the normal stresses on

Â two orthogonal planes is going to be invariant, or constant.

Â And so that's where we will leave off this time.

Â Some important relationships.

Â And we'll continue on next time.

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Â